Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
int dx8[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dy8[] = {1, -1, -1, 0, 1, -1, 0, 1};
const int M = 100005;
vector<int> adj[M], vec[M];
int par[M], sp[M][22], lvl[M], id, a[M], root[M], n, m, q;
int avail;
struct node {
  int l, r, val;
  node() { l = r = val = 0; }
} sum[M * 80];
int update(int PreNode, int l, int r, int L, int val) {
  int NewNode = ++avail;
  if (l == r) {
    sum[NewNode].val = val;
    return NewNode;
  }
  int mid = (l + r) / 2;
  if (L <= mid) {
    sum[NewNode].r = sum[PreNode].r;
    sum[NewNode].l = update(sum[PreNode].l, l, mid, L, val);
  } else {
    sum[NewNode].l = sum[PreNode].l;
    sum[NewNode].r = update(sum[PreNode].r, mid + 1, r, L, val);
  }
  sum[NewNode].val = sum[sum[NewNode].l].val + sum[sum[NewNode].r].val;
  return NewNode;
}
long long query(int x, int y, int lca, int plca, int l, int r, int k) {
  if (l == r) {
    return l;
  }
  int mid = (l + r) / 2;
  int total = sum[sum[x].l].val + sum[sum[y].l].val - sum[sum[lca].l].val -
              sum[sum[plca].l].val;
  if (total >= k) {
    return query(sum[x].l, sum[y].l, sum[lca].l, sum[plca].l, l, mid, k);
  } else {
    return query(sum[x].r, sum[y].r, sum[lca].r, sum[plca].r, mid + 1, r,
                 k - total);
  }
}
void dfs(int p, int x, int lev) {
  root[x] = root[p];
  for (auto z : vec[x]) {
    root[x] = update(root[x], 1, m, z, 1);
  }
  par[x] = p;
  lvl[x] = lev;
  for (auto y : adj[x])
    if (p != y) {
      dfs(x, y, lev + 1);
    }
}
void build(int n) {
  for (int i = 1; i <= n; i++) {
    sp[i][0] = par[i];
  }
  for (int i = 1; (1 << i) <= n; i++) {
    for (int j = 1; j <= n; j++) {
      sp[j][i] = sp[sp[j][i - 1]][i - 1];
    }
  }
}
int lca_of(int u, int v) {
  if (lvl[u] > lvl[v]) swap(u, v);
  for (int i = 17; i >= 0; i--) {
    if (!sp[v][i]) continue;
    if (lvl[sp[v][i]] >= lvl[u]) {
      v = sp[v][i];
    }
  }
  if (u == v) return u;
  for (int i = 17; i >= 0; i--) {
    if (sp[u][i] == sp[v][i]) continue;
    u = sp[u][i];
    v = sp[v][i];
  }
  return par[u];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> q;
  for (int i = 0; i < (n - 1); i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  for (int i = 1; i <= m; i++) {
    int a;
    cin >> a;
    vec[a].push_back(i);
  }
  dfs(0, 1, 1);
  build(n);
  while (q--) {
    int x, y, z;
    cin >> x >> y >> z;
    int lca = lca_of(x, y);
    int plca = par[lca];
    int total = sum[root[x]].val + sum[root[y]].val - sum[root[lca]].val -
                sum[root[plca]].val;
    z = min(z, total);
    cout << z;
    for (int i = 1; i <= z; i++) {
      cout << " " << query(root[x], root[y], root[lca], root[plca], 1, m, i);
    }
    cout << "\n";
  }
  return 0;
}